《2025年同步练习册河北教育出版社九年级数学上册冀教版》

22. (本小题满分$11$分)
在$Rt\triangle ABC$中,$\angle ACB = 90^{\circ}$,$AC = 6 cm$,$BC = 8 cm$,动点$P从点B$出发,在$BA边上以5 cm/s的速度向点A$匀速运动,同时动点$Q从点C$出发,在$CB边上以4 cm/s的速度向点B$匀速运动,运动时间为$t(s)(0\lt t\lt2)$.
(1)如图①,连接$PQ$,若$\triangle BPQ与\triangle ABC$相似,求$t$的值.
(2)如图②,连接$AQ$,$CP$,若$AQ\perp CP$,求$t$的值.

答案：
解:
(1)由题意,得BP = 5t,QC = 4t,BA = $\sqrt{AC^{2}+BC^{2}}$ = 10(cm).①当△BPQ∽△BAC时,$\dfrac{BP}{BA}=\dfrac{BQ}{BC}$,
∴$\dfrac{5t}{10}=\dfrac{8 - 4t}{8}$,解得t = 1.②当△BPQ∽△BCA时,$\dfrac{BP}{BC}=\dfrac{BQ}{BA}$,
∴$\dfrac{5t}{8}=\dfrac{8 - 4t}{10}$,解得t = $\dfrac{32}{41}$.
∴当t = 1s或$\dfrac{32}{41}$s时,△BPQ与△ABC相似.
(2)如图,过点P作PM⊥BC于点M,AQ,CP交于点N,则△BPM∽△BAC,
∴$\dfrac{BP}{BA}=\dfrac{PM}{AC}=\dfrac{BM}{BC}$,
∴易得BP = 5t,MP = 3t,BM = 4t,CM = 8 - 4t.
∵∠NAC + ∠NCA = 90°,∠PCM + ∠NCA = 90°,
∴∠NAC = ∠PCM.又
∵∠ACQ = ∠CMP = 90°,
∴△ACQ∽△CMP,
∴$\dfrac{AC}{CM}=\dfrac{CQ}{MP}$,
∴$\dfrac{6}{8 - 4t}=\dfrac{4t}{3t}$,解得t = $\dfrac{7}{8}$.
∴当t = $\dfrac{7}{8}$s时,AQ⊥CP.